This would be a permutation question.
We have 4 choices, 0, 3, 7, 8
As we wish to make a 4-digit number abcd.

For d, the units, we have 2 choices, 3 or 7.
And for each of the other digits, a,b,c we have 4 choices, 0,3,7,8.

Hence the our total number of choices are 4x4x4x2

You were on the right track, however a 4 digit number cant start with a 0.
That is:
Say, we wish to make a 4-digit number abcd.
We want a 4-digit number, so acannot be 0. So we have only 3 choices for a.

Hence the our total number of choices are 3x4x4x2

Originally Posted by JSB1917

I'm pretty sure it would just be 4! (that's factorial not an exclamation).
Oh and 4! = 24.

Hey JSB1917,
Why do you think it is 4!? that would be counting all permutations of the letters of the string 0378. But we need the total number of "odd" "numbers" in such strings. Note that 0378 itself is even

Thank you for all of your posts and help.
Today in math class we got to this exercise again, tho a bit different example. By the way, turns out that I forgot to mention a detail about that exercise I gave in my initial post - the numbers can't be repeated.
So anyways here's what exercise we solved today:
How many even, 4 digit numbers, without repeating the digits can you make out of 0, 2, 7 and 9?
Here's what the teacher wrote on the board for this one:
So then she wrote all the possible combinations, which there were 18, and then we chose all numbers that were even, and we got only 10 numbers. So the answer is that you can make 10 even, 4 digit numbers, without repeating the digits out of 0, 2, 7 and 9 .